To determine To solve: The ODE y ″ + 0.54 y ′ + ( 0.0729 + π ) y = 0 for the given initial values and check whether the answer satisfies the ODE also the initial conditions. Explanation Case (i): If the auxiliary equation has distinct real roots λ 1 , and λ 2 of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e λ 1 x , e λ 2 x and the general solution of y ″ + a y ′ + b y = 0 is y = c 1 e λ 1 x + c 2 e λ 2 x . Case (ii): If the auxiliary equation has real double root λ = − a of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e − a x , x e − a x and the general solution of y ″ + a y ′ + b y = 0 is y = ( c 1 + c 2 ) e − a x . Case (iii): If the auxiliary equation has complex conjugate root λ 1 = a + i ω , λ 2 = a − i ω of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e a x cos ω x , e a x sin ω x and the general solution of y ″ + a y ′ + b y = 0 is y = e − a x ( A cos ω x + B sin ω x ) . Calculation: The given ODE is y ″ + 0.54 y ′ + ( 0.0729 + π ) y = 0 (1) Compute the auxiliary equation of the above ODE as follows. Let y = e λ x . That implies, the first and second derivative of y is y ′ = λ e λ x and y ″ = λ 2 e λ x , respectively. Substitute y ″ = λ 2 e λ x and y = e λ x in (1) and simplify further. λ 2 e λ x + 0.54 λ e λ x + ( 0.0729 + π ) e λ x = 0 e λ x ( λ 2 + 0.54 λ + ( 0.0729 + π ) ) = 0 λ 2 + 0.54 λ + ( 0.0729 + π ) = 0 That is, the characteristic equation is λ 2 + 0.54 λ + ( 0.0729 + π ) = 0 . Compute the roots of the above characteristic equation as follows. Substitute a = 1, b = 0.54 and c = 0.0729 + π in λ = − b ± b 2 − 4 a c 2 a . λ = − 0.54 ± ( 0.54 ) 2 − 4 ( 1 ) ( 0.0729 + π ) 2 ( 1 ) = − 0.54 ± 0.2916 − 0.2916 − 4 π 2 = − 0.54 ± 2 i π 2 = − 0.27 + i π or − 0.37 − i π From the above calculation, it is observed that y ″ + 0.54 y ′ + ( 0.0729 + π ) y = 0 has complex conjugate roots. Then by case (iii), the basis of y ″ + 0.54 y ′ + ( 0.0729 + π ) y = 0 is e − 0.27 x cos ( π x ) , e − 0.27 x sin ( π x ) . The general solution of y ″ + 0.54 y ′ + ( 0.0729 + π ) y = 0 is y = e − 0.27 x ( c 1 cos ( π x ) + c 2 sin ( π x ) ) . Now check the correctness of the above solution as follows. Compute the second derivative of y as follows. y ′ = { e − 0.27 x ( − 0.27 ) ( c 1 cos ( π x ) + c 2 sin ( π x ) ) + e − 0.27 x ( − c 1 sin ( π x ) ( π ) + c 2 cos ( π x ) ( π ) ) } = { − 0.27 e − 0.27 x ( c 1 cos ( π x ) + c 2 sin ( π x ) ) + e − 0.27 x ( − c 1 sin ( π x ) ( π ) + c 2 cos ( π x ) ( π ) ) } y ″ = { − 0.27 e − 0.27 x ( − 0.27 ) ( c 1 cos ( π x ) + c 2 sin ( π x ) ) − 0.27 e − 0.27 x ( − c 1 sin ( π x ) ( π ) + c 2 cos ( π x ) ( π ) ) + e − 0.27 x ( − 0.27 ) ( − π c 1 sin ( π x ) + π c 2 cos ( π x ) ) + e − 0.27 x ( − π c 1 cos ( π x ) − π c 2 sin ( π x ) ) } = { 0.0729 c 1 e − 0.27 x cos ( π x ) + 0.0729 c 2 e − 0.27 x sin ( π x ) + 0

Custom Kreyszig: Advanced Engineering Mathematics

10th Edition

ISBN: 9781119166856

Author: Kreyszig

Publisher: JOHN WILEY+SONS INC.CUSTOM

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Chapter 2.2, Problem 29P

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To solve: The ODE y″+0.54y′+(0.0729+π)y=0 for the given initial values and check whether the answer satisfies the ODE also the initial conditions.

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